Can you help me find a root for $c$ in the equation below? $$ce^{-c}-{10\over5}(1-e^{-c})^2=0$$
By expanding this I got, $$ce^{-c}-2 + 4 e^{-c}-2e^{-2c}=0$$ now grouping, $$(c+4)e^{-c}-2-2e^{-2c}=0 \tag{1} $$ Let $$ e^{-c} = x$$ then $$ c = -\log x$$ Substituting these values in (1), $$x^2 + (\log x - 4) x +1 = 0$$
Now if I apply formula to find the root of quadratic equation, $(\log x -4)$ term is coming inside the square root and making it complex to find the root. Am I proceeding right? Is there any other way to find the root of this equation?
If I plot a graph for this the curve is cutting x axis at $0.49$, which is one of the root. How to arrive at this mathematically?
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Working with $c$ and rearranging the equation, we look for the non-trivial zero of function $$f(c)=2 e^{2 c}- (c+4)\,e^c+2$$for which $$f'(c)= e^{ c}\, \left(4 e^c-c-5\right)\qquad \text{and}\qquad f''(c)=e^{ c}\, \left(8 e^c-c-6\right)$$
The first derivative cancels at $$c_*=-5-W_{-1}\left(-\frac{4}{e^5}\right)$$ where $W(.)$ is Lambert function. The second derivative test reveals that this is a minimum.
Making a series expansion around $c_*$
$$f(c)=f(c_*)+\frac 12 f'(c_*) (c-c_*)^2 + O((c-c_*)^3)$$ Then an estimate $$c_0=c_*+\sqrt{-2\frac{f(c_*)}{f''(c_*)}}$$
Converted to decimal, this gives $c_0=0.516463$ while the solution, given by Newton method, is $c=0.490110$.
Now, expanding $f(c)$ as a series around $c=\frac 12$ and then series reversion $$c=\frac{1}{2}+t+\frac{\left(13-16 \sqrt{e}\right) }{2 \left(8 \sqrt{e}-11\right)}t^3+O\left(t^{4}\right)\qquad \text{with}\qquad t=\frac{\left(4-9 \sqrt{e}+4 e\right)+2 f(c) }{11 \sqrt{e}-8 e }$$ Since we want $f(c)=0$ then, numerically $c=0.490113$.